Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $t = \dfrac{-9k - 45}{k^2 + 6k - 16} \times \dfrac{-k + 2}{k + 5} $
Answer: First factor the quadratic. $t = \dfrac{-9k - 45}{(k - 2)(k + 8)} \times \dfrac{-k + 2}{k + 5} $ Then factor out any other terms. $t = \dfrac{-9(k + 5)}{(k - 2)(k + 8)} \times \dfrac{-(k - 2)}{k + 5} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ -9(k + 5) \times -(k - 2) } { (k - 2)(k + 8) \times (k + 5) } $ $t = \dfrac{ 9(k + 5)(k - 2)}{ (k - 2)(k + 8)(k + 5)} $ Notice that $(k + 5)$ and $(k - 2)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ 9(k + 5)\cancel{(k - 2)}}{ \cancel{(k - 2)}(k + 8)(k + 5)} $ We are dividing by $k - 2$ , so $k - 2 \neq 0$ Therefore, $k \neq 2$ $t = \dfrac{ 9\cancel{(k + 5)}\cancel{(k - 2)}}{ \cancel{(k - 2)}(k + 8)\cancel{(k + 5)}} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $t = \dfrac{9}{k + 8} ; \space k \neq 2 ; \space k \neq -5 $